for some , where are the Pell numbers. and are the half-companion Pell numbers, and they are given by for each . So we have to find all such that

.

On the other hand, if , then for some integer . Then . Hence implies

.

Now , so for some integers . Clearly , so

and .

So is a near isosceles Pythagorean triple, i.e.

.

Hence

.

The two boxed cases are what only remain to check, and they show why this actually is a difficult problem! (Refer to the articles about Fibonacci numbers in the ‘Notes and Articles’ tab above to see how little information we have on divisors of sequences of this kind.) Nevertheless, I’ll leave it here and maybe come back to it another time.

Let and be the roots of . We can factorise as . If is rational, it must be an integer and then it is easy to find all solutions. So let us consider the more interesting case where is not rational (i.e. ).

Now a solution in the form to is a unit in the ring. For , denote its conjugate by . Then we have the multiplicative norm